'''
Description: 
version: 
Company: 
Author: Qing Bei
Date: 2020-08-22 21:08:09
LastEditors: Qing Bei
LastEditTime: 2020-08-25 15:49:01
'''
# lst = [['x',1,1,1], # a
#        [1,1,'x',1], # b
#        [1,1,1,'x'], # c
#        [1,1,'x',1]  # d
#        ]

# 方法1-暴力法
    # 行遍历
    # flg = False
    # for i in range(len(lst)):
    #     # 列遍历
    #     for j in range(len(lst)):
    #         # 该判断是假设有两个人的说法冲突 之后找第三个人辩证
    #         if lst[i][j] is lst[j][i] and lst[i][j]=='x' and i != j and flg ==False:  # 第三个人的辩证判断条件
    #                 if lst[i][j] in [lst[i][2] for i in range(len(lst))] :
    #                     print(chr(i+97))
    #                     flg = True
    #                     break
    #                 else:
    #                     print(chr(j+97))
    #                     flg = True
    #                     break
    
# 方法2 ——_——
for x in ["a","b","c","d"]:
    sum1 = ("a"!=x) + ("c"==x) + ("d"==x) + ("d"!=x) # 执行判断 每次对一个人的说法代入结论 结果为3说明有三个人说的是对的 
    if sum1 == 3:
        print(x)
# 方法3 和方法2一样
for x in ["a","b","c","d"]:
    if (("a"!=x) + ("c"==x) + ("d"==x) + ("d"!=x)==3):
        print(x)